Some properties of meromorphically multivalent functions

Abstract

By using the method of differential subordinations, we derive certain properties of meromorphically multivalent functions.

2010 Mathematics Subject Classification: 30C45; 30C55.

1 Introduction

Let Σ(p) denotes the class of meromorphically multivalent functions f(z) of the form

$$f\left(z\right)=z^{-p}+\sum _{k=1}^{\infty }{a}_{k-p}{z}^{k-p}\left(p\in N=\left\{1,2,3,\dots \right\}\right),$$

(1.1)

which are analytic in the punctured unit disk

$$U^{*}=\left\{z:z\in C\mathsf{\text{and}}\mathsf{\text{0}}<\left|z\right|<1\right\}=U\backslash \left\{0\right\}.$$

Let f(z) and g(z) be analytic in U. Then, we say that f(z) is subordinate to g(z) in U, written f(z) ≺ g(z), if there exists an analytic function w(z) in U, such that |w(z)| ≤ |z| and f(z) = g(w(z)) (z ∈ U). If g(z) is univalent in U, then the subordination f(z) ≺ g(z) is equivalent to f(0) = g(0) and f(U) ⊂ g(U).

Let p(z) = 1 + p₁z + ... be analytic in U. Then for -1 ≤ B < A ≤ 1, it is clear that

$$p\left(z\right)\prec \frac{1+Az}{1+Bz}\left(z\in U\right)$$

(1.2)

if and only if

$$\left|p\left(z\right)-\frac{1-AB}{1-B^2}\right|<\frac{A-B}{1-B^2}\left(-1<B<A\le 1;z\in U\right)$$

(1.3)

and

$$\mathsf{\text{Re}}p\left(z\right)>\frac{1-A}{2}\left(B=-1;z\in U\right).$$

(1.4)

Recently, several authors (see, e.g., [1–7]) considered some interesting properties of meromorphically multivalent functions. In the present article, we aim at proving some subordination properties for the class Σ(p).

To derive our results, we need the following lemmas.

Lemma 1 (see [8]. Let h(z) be analytic and starlike univalent in U with h(0) = 0. If g(z) is analytic in U and zg'(z) ≺ h(z), then

$$g\left(z\right)\prec g\left(0\right)+\underset{0}{\overset{z}{\int }}\frac{h\left(t\right)}{t}dt.$$

Lemma 2 (see [9]. Let p(z) be analytic and nonconstant in U with p(0) = 1. If 0 < | z₀ | < 1 and $\mathsf{\text{Re}}p\left(z_0\right)=\underset{\left|z\right|\le \left|z_0\right|}{\text{min}}\mathsf{\text{Re}}p\left(z\right)$, then

$$z_0{p}^{\prime }\left(z_0\right)\le -\frac{{\left|1-p\left(z_0\right)\right|}^2}{2\left(1-\mathsf{\text{Re}}p\left(z_0\right)\right)}.$$

2 Main results

Our first result is contained in the following.

Theorem 1. Let $\alpha \in \left(0,\frac{1}{2}\right]$ and β ∈ (0,1). If f(z) ∈ Σ(p) satisfies f(z) ≠ 0 (z ∈ U*) and

$$\left|\frac{z^{-p}}{f\left(z\right)}\left(\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}+p\right)\right|<\delta \left(z\in U\right),$$

(2.1)

where δ is the minimum positive root of the equation

$$\alpha \text{sin}\left(\frac{\pi \beta }{2}\right)x^2-x+\left(1-\alpha \right)\text{sin}\left(\frac{\pi \beta }{2}\right)=0,$$

(2.2)

then

$$\left|\text{arg}\left(\frac{f\left(z\right)}{z^{-p}}-\alpha \right)\right|<\frac{\pi }{2}\beta \left(z\in U\right).$$

(2.3)

The bound β is the best possible for each $\alpha \in \left(0,\frac{1}{2}\right]$.

Proof. Let

$$g\left(x\right)=\alpha \text{sin}\left(\frac{\pi \beta }{2}\right)x^2-x+\left(1-\alpha \right)\text{sin}\left(\frac{\pi \beta }{2}\right).$$

(2.4)

We can see that the Equation (2.2) has two positive roots. Since g(0) > 0 and g(1) < 0, we have

$$0<\frac{\alpha }{1-\alpha }\delta \le \delta <1.$$

(2.5)

Put

$$\frac{f\left(z\right)}{z^{-p}}=\alpha +\left(1-\alpha \right)p\left(z\right).$$

(2.6)

Then from the assumption of the theorem, we see that p(z) is analytic in U with p(0) = 1 and α + (1 - α)p(z) ≠ 0 for all z ∈ U. Taking the logarithmic differentiations in both sides of (2.6), we get

$$\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}+p=\frac{\left(1-\alpha \right)z{p}^{\prime }\left(z\right)}{\alpha +\left(1-\alpha \right)p\left(z\right)}$$

(2.7)

and

$$\frac{z^{-p}}{f\left(z\right)}\left(\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}+p\right)=\frac{\left(1-\alpha \right)z{p}^{\prime }\left(z\right)}{{\left(\alpha +\left(1-\alpha \right)p\left(z\right)\right)}^2}$$

(2.8)

for all z ∈ U. Thus the inequality (2.1) is equivalent to

$$\frac{\left(1-\alpha \right)z{p}^{\prime }\left(z\right)}{{\left(\alpha +\left(1-\alpha \right)p\left(z\right)\right)}^2}\prec \delta z.$$

(2.9)

By using Lemma 1, (2.9) leads to

$$\underset{0}{\overset{z}{\int }}\frac{\left(1-\alpha \right)p^{\prime }\left(t\right)}{{\left(\alpha +\left(1-\alpha \right)p\left(t\right)\right)}^2}dt\prec \delta z$$

or to

$$1-\frac{1}{\alpha +\left(1-a\right)p\left(z\right)}\prec \delta z.$$

(2.10)

In view of (2.5), (2.10) can be written as

$$p\left(z\right)\prec \frac{1+\frac{\alpha }{1-\alpha }\delta z}{1-\delta z}.$$

(2.11)

Now by taking $A=\frac{\alpha }{1-\alpha }\delta$ and B = -δ in (1.2) and (1.3), we have

$$\begin{array}{ll}\hfill \left|\text{arg}\left(\frac{f\left(z\right)}{z^{-p}}-\alpha \right)\right|& =\left|\text{arg}p\left(z\right)\right|\\ <\text{arcsin}\left(\frac{\delta }{1-\alpha +\alpha {\delta }^2}\right)\\ =\frac{\pi }{2}\beta \end{array}$$

for all z ∈ U because of g(δ) = 0. This proves (2.3).

Next, we consider the function f(z) defined by

$$f\left(z\right)=\frac{z^{-p}}{1-{\delta }_z}\left(z\in U^{*}\right).$$

It is easy to see that

$$\left|\frac{z^{-p}}{f\left(z\right)}\left(\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}+p\right)\right|=\left|\delta z\right|<\delta \left(z\in U\right).$$

Since

$$\frac{f\left(z\right)}{z^{-p}}-\alpha =\left(1-\alpha \right)\frac{1+\frac{\alpha }{1-\alpha }\delta z}{1-\delta z},$$

it follows from (1.3) that

$$\underset{z\in U}{\text{sup}}\left|\text{arg}\left(\frac{f\left(z\right)}{z^{-p}}-\alpha \right)\right|=\text{arcsin}\left(\frac{\delta }{1-\alpha +\alpha {\delta }^2}\right)=\frac{\pi }{2}\beta .$$

Hence, we conclude that the bound β is the best possible for each $\alpha \in \left(0,\frac{1}{2}\right]$.

Next, we derive the following.

Theorem 2. If f(z) ∈ Σ(p) satisfies f(z) ≠ 0 (z ∈ U*) and

$$\mathsf{\text{Re}}\left\{\frac{z^{-p}}{f\left(z\right)}\left(\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}+p\right)\right\}<\gamma \left(z\in U\right),$$

(2.12)

where

$$0<\gamma <\frac{1}{2\text{log}2},$$

(2.13)

then

$$\mathsf{\text{Re}}\frac{z^{-p}}{f\left(z\right)}>1-2\gamma \text{log}2\left(z\in U\right).$$

(2.14)

The bound in (2.14) is sharp.

Proof. Let

$$p\left(z\right)=\frac{f\left(z\right)}{z^{-p}}.$$

(2.15)

Then p(z) is analytic in U with p(0) = 1 and p(z) ≠ 0 for z ∈ U. In view of (2.15) and (2.12), we have

$$1-\frac{z{p}^{\prime }\left(z\right)}{\gamma p^2\left(z\right)}\prec \frac{1+z}{1-z},$$

i.e.,

$$z{\left(\frac{1}{p\left(z\right)}\right)}^{\prime }\prec \frac{2\gamma z}{1-z}.$$

Now by using Lemma 1, we obtain

$$\frac{1}{p\left(z\right)}\prec 1-2\gamma \text{log}\left(1-z\right).$$

(2.16)

Since the function 1 - 2γ log(1 - z) is convex univalent in U and

$$\mathsf{\text{Re}}\left(1-2\gamma \text{log}\left(1-z\right)\right)>1-2\gamma \text{log}2\left(z\in U\right),$$

from (2.16), we get the inequality (2.14).

To show that the bound in (2.14) cannot be increased, we consider

$$f\left(z\right)=\frac{z^{-p}}{1-2\gamma \text{log}\left(1-z\right)}\left(z\in U^{*}\right).$$

It is easy to verify that the function f(z) satisfies the inequality (2.12). On the other hand, we have

$$\mathsf{\text{Re}}\frac{z^{-p}}{f\left(z\right)}\to 1-2\gamma \text{log}2$$

as z → -1. Now the proof of the theorem is complete.

Finally, we discuss the following theorem.

Theorem 3. Let f(z) ∈ Σ(p) with f(z) ≠ 0 (z ∈ U*). If

$$\left|\mathsf{\text{Im}}\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\left(\frac{f\left(z\right)}{z^{-p}}-\lambda \right)\right\}\right|<\sqrt{\lambda \left(\lambda +2p\right)}\left(z\in U\right)$$

(2.17)

for some λ(λ > 0), then

$$\mathsf{\text{Re}}\frac{f\left(z\right)}{z^{-p}}>0\left(z\in U\right).$$

(2.18)

Proof. Let us define the analytic function p(z) in U by

$$\frac{f\left(z\right)}{z^{-p}}=p\left(z\right).$$

Then p(0) = 1, p(z) ≠ 0 (z ∈ U) and

$$\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\left(\frac{f\left(z\right)}{z^{-p}}-\lambda \right)=\left(p\left(z\right)-\lambda \right)\left(\frac{z{p}^{\prime }\left(z\right)}{p\left(z\right)}-p\right)\left(z\in U\right).$$

(2.19)

Suppose that there exists a point z₀(0 < | z₀ | < 1) such that

$$\mathsf{\text{Re}}p\left(z\right)>0\left(\left|z\right|<\left|z_0\right|\right)\mathsf{\text{and}}p\left(z_0\right)=i\beta ,$$

(2.20)

where β is real and β ≠ 0. Then, applying Lemma 2, we get

$$z_0{p}^{\prime }\left(z_0\right)\le -\frac{1+{\beta }^2}{2}.$$

(2.21)

Thus it follows from (2.19), (2.20), and (2.21) that

$$I_0=\mathsf{\text{Im}}\left\{\frac{z_0{f}^{\prime }\left(z_0\right)}{f\left(z_0\right)}\left(\frac{f\left(z_0\right)}{z_0^{-p}}-\lambda \right)\right\}=-p\beta +\frac{\lambda }{\beta }{z}_0{p}^{\prime }\left(z_0\right).$$

(2.22)

In view of λ > 0, from (2.21) and (2.22) we obtain

$$I_0\ge -\frac{\lambda +\left(\lambda +2p\right){\beta }^2}{2\beta }\ge \sqrt{\lambda \left(\lambda +2p\right)}\left(\beta <0\right)$$

(2.23)

and

$$I_0\le -\frac{\lambda +\left(\lambda +2p\right){\beta }^2}{2\beta }\le -\sqrt{\lambda \left(\lambda +2p\right)}\left(\beta >0\right).$$

(2.24)

But both (2.23) and (2.24) contradict the assumption (2.17). Therefore, we have Rep(z) > 0 for all z ∈ U. This shows that (2.18) holds true.